Problem: Solve for $x$ and $y$ using substitution. ${-2x+3y = 9}$ ${x = 2y-5}$
Since $x$ has already been solved for, substitute $2y-5$ for $x$ in the first equation. ${-2}{(2y-5)}{+ 3y = 9}$ Simplify and solve for $y$ $-4y+10 + 3y = 9$ $-y+10 = 9$ $-y+10{-10} = 9{-10}$ $-y = -1$ $\dfrac{-y}{{-1}} = \dfrac{-1}{{-1}}$ ${y = 1}$ Now that you know ${y = 1}$ , plug it back into $\thinspace {x = 2y-5}\thinspace$ to find $x$ ${x = 2}{(1)}{ - 5}$ $x = 2 - 5$ ${x = -3}$ You can also plug ${y = 1}$ into $\thinspace {-2x+3y = 9}\thinspace$ and get the same answer for $x$ : ${-2x + 3}{(1)}{= 9}$ ${x = -3}$